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 Post subject: Strictly Convex FunctionsPosted: Tue, 31 Jan 2012 21:20:33 UTC
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 Post subject: Re: Strictly Convex FunctionsPosted: Tue, 31 Jan 2012 21:21:58 UTC
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Joined: Thu, 19 Feb 2009 05:25:19 UTC
Posts: 114
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).

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 Post subject: Re: Strictly Convex FunctionsPosted: Tue, 31 Jan 2012 21:23:27 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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kingwinner wrote:

What is your little circle thingy?

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 Post subject: Re: Strictly Convex FunctionsPosted: Tue, 31 Jan 2012 23:19:51 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).

Think about the line segment . What must f be if you have ?

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 Post subject: Re: Strictly Convex FunctionsPosted: Wed, 1 Feb 2012 00:54:57 UTC
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Joined: Thu, 19 Feb 2009 05:25:19 UTC
Posts: 114
outermeasure wrote:
kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).

Think about the line segment . What must f be if you have ?

(The little circle thingy is dot product.)

Let x y.
Suppose f(y) = f(x) + grad f(x) o (y-x).
By strict convexity, for all 0<s<1,
f[(1-s)x+sy] < (1-s)f(x) + s f(y)
f[x+s(y-x)]< (1-s)f(x) + s [f(x) + grad f(x) o (y-x)]= f(x) + s [grad f(x) o (y-x)]

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 Post subject: Re: Strictly Convex FunctionsPosted: Wed, 1 Feb 2012 01:20:30 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6066
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
kingwinner wrote:
outermeasure wrote:
kingwinner wrote:
So I need to show that f(y) = f(x) + grad f(x) o (y-x) => x=y. But I can't think of a way of showing it...
What kind of manipluations do I have to do with f(y) = f(x) + grad f(x) o (y-x)? I have no idea what to do next after assuming f(y) = f(x) + grad f(x) o (y-x).

Think about the line segment . What must f be if you have ?

(The little circle thingy is dot product.)

Let x y.
Suppose f(y) = f(x) + grad f(x) o (y-x).
By strict convexity, for all 0<s<1,
f[(1-s)x+sy] < (1-s)f(x) + s f(y)
f[x+s(y-x)]< (1-s)f(x) + s [f(x) + grad f(x) o (y-x)]= f(x) + s [grad f(x) o (y-x)]

So you reduce that to univariate case. What is the relationship between secants and tangents there? What does that tell you?

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